Why Buy and Sell Stocks All Variants in DSA C?

The Big Idea

Discover how simple rules can turn confusing stock prices into clear profit opportunities!

The Scenario

Imagine you want to make money by buying and selling stocks. You try to track prices on paper and decide when to buy or sell manually.

You write down prices every day and try to guess the best days to buy low and sell high.

The Problem

This manual way is slow and confusing. You might miss the best days or sell too early.

It is hard to remember all prices and compare them quickly, especially if prices change every minute.

The Solution

Using algorithms to buy and sell stocks helps you find the best days to buy and sell automatically.

These algorithms check all prices fast and tell you when to act to get the most profit.

Before vs After

✗ Before

int maxProfit(int prices[], int n) {
    int max_profit = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (prices[j] > prices[i]) {
                int profit = prices[j] - prices[i];
                if (profit > max_profit) max_profit = profit;
            }
        }
    }
    return max_profit;
}

✓ After

int maxProfit(int prices[], int n) {
    int min_price = prices[0];
    int max_profit = 0;
    for (int i = 1; i < n; i++) {
        if (prices[i] < min_price) min_price = prices[i];
        else if (prices[i] - min_price > max_profit) max_profit = prices[i] - min_price;
    }
    return max_profit;
}

What It Enables

It enables you to quickly find the best buy and sell points to maximize profit without guessing.

Real Life Example

Stock traders use these algorithms to decide when to buy or sell shares automatically, saving time and increasing earnings.

Key Takeaways

Manual tracking of stock prices is slow and error-prone.

Algorithms find the best buy/sell days quickly and accurately.

This helps maximize profit with less effort and mistakes.

Practice

1. What is the main goal when solving the Buy and Sell Stocks problem variants?

easy

A. Maximize the profit by choosing the best days to buy and sell stocks

B. Minimize the number of transactions regardless of profit

C. Find the day with the highest stock price only

D. Calculate the average stock price over all days

Solution

1. Step 1: Understand the problem objective

   The problem asks to maximize profit by buying and selling stocks on certain days.

2. Step 2: Identify the key action

   Choosing the best days to buy low and sell high maximizes profit.

3. Final Answer:

   Maximize the profit by choosing the best days to buy and sell stocks -> Option A

4. Quick Check:

   Goal = Maximize profit [OK]

Hint: Focus on maximizing profit, not just counting days [OK]

Common Mistakes:

• Confusing profit maximization with minimizing transactions
• Ignoring the selling day after buying
• Thinking only about highest price day

2. Which of the following C code snippets correctly initializes the variable to track maximum profit as zero?

easy

A. int maxProfit = 0;

B. int maxProfit = -1;

C. int maxProfit = NULL;

D. int maxProfit = '0';

Solution

1. Step 1: Check variable initialization syntax

   In C, integer variables are initialized with integer values, zero is valid.

2. Step 2: Validate each option

   int maxProfit = 0; uses int maxProfit = 0; which is correct. int maxProfit = -1; uses -1 which is valid but not typical for profit initialization. int maxProfit = NULL; uses NULL which is for pointers, not int. int maxProfit = '0'; uses character '0' which is ASCII 48, not zero.

3. Final Answer:

   int maxProfit = 0; -> Option A

4. Quick Check:

   Correct int initialization = int maxProfit = 0; [OK]

Hint: Use zero integer for profit start, not NULL or char [OK]

Common Mistakes:

• Using NULL for integer variables
• Initializing with character '0' instead of number 0
• Starting profit with negative values unnecessarily

3. What will be the output of the following C code that finds max profit with one transaction?

int prices[] = {7, 1, 5, 3, 6, 4};
int maxProfit = 0, minPrice = prices[0];
for (int i = 1; i < 6; i++) {
    if (prices[i] < minPrice) minPrice = prices[i];
    else if (prices[i] - minPrice > maxProfit) maxProfit = prices[i] - minPrice;
}
printf("%d", maxProfit);

medium

A. 4

B. 5

C. 6

D. 7

Solution

1. Step 1: Trace the loop to find minPrice and maxProfit

   Start minPrice=7, maxProfit=0. Iterations:
   i=1: price=1 < 7, minPrice=1
   i=2: price=5, profit=5-1=4 > 0, maxProfit=4
   i=3: price=3, profit=3-1=2 < 4, no change
   i=4: price=6, profit=6-1=5 > 4, maxProfit=5
   i=5: price=4, profit=4-1=3 < 5, no change

2. Step 2: Final maxProfit after loop

   maxProfit is 5 after all iterations.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Max profit = 5 [OK]

Hint: Track min price and update max profit on each day [OK]

Common Mistakes:

• Not updating minPrice correctly
• Calculating profit before updating minPrice
• Confusing maxProfit with max price

4. Identify the bug in this C code snippet that tries to calculate max profit with unlimited transactions:

int maxProfit = 0;
int prices[] = {1, 2, 3, 4, 5};
for (int i = 0; i < 5; i++) {
    if (prices[i] > prices[i-1]) {
        maxProfit += prices[i] - prices[i-1];
    }
}
printf("%d", maxProfit);

medium

A. The condition should be prices[i] < prices[i-1]

B. maxProfit should be initialized to -1

C. Loop starts at i=0 causing out-of-bounds access at prices[i-1]

D. The array size is incorrect

Solution

1. Step 1: Analyze loop index usage

   The loop starts at i=0, so prices[i-1] is prices[-1], which is invalid memory access.

2. Step 2: Correct the loop start

   Loop should start at i=1 to safely access prices[i-1].

3. Final Answer:

   Loop starts at i=0 causing out-of-bounds access at prices[i-1] -> Option C

4. Quick Check:

   Index error at i=0 [OK]

Hint: Start loop from 1 when accessing previous element [OK]

Common Mistakes:

• Starting loop at 0 when accessing i-1
• Incorrect comparison operator in if condition
• Ignoring array bounds

5. Given the prices array {3, 3, 5, 0, 0, 3, 1, 4}, what is the maximum profit with at most two transactions?

hard

A. 8

B. 6

C. 5

D. 7

Solution

1. Step 1: Understand two transactions limit

   We can buy and sell stocks at most twice to maximize profit.

2. Step 2: Calculate max profit for two transactions

   First transaction: buy at 3 (day 0), sell at 5 (day 2), profit = 2.
   Second transaction: buy at 0 (day 3), sell at 4 (day 7), profit = 4.
   Total profit = 2 + 4 = 6.

3. Step 3: Check for better transactions

   Alternatively, buy at 0 (day 3), sell at 3 (day 5), profit = 3.
   Then buy at 1 (day 6), sell at 4 (day 7), profit = 3.
   Total profit = 3 + 3 = 6.
   But better is buy at 0 (day 3), sell at 5 (day 2) is invalid (sell before buy). So best is 6 or 7 or 8?
   Actually, the maximum profit with two transactions is 7:
   Buy at 0 (day 3), sell at 3 (day 5), profit = 3.
   Buy at 1 (day 6), sell at 4 (day 7), profit = 3.
   Total profit = 6.
   But if we consider buy at 3 (day 0), sell at 5 (day 2), profit = 2.
   Buy at 0 (day 3), sell at 4 (day 7), profit = 4.
   Total profit = 6.
   However, the maximum profit is 7:
   Buy at 0 (day 3), sell at 5 (day 2) invalid.
   Buy at 0 (day 3), sell at 4 (day 7), profit = 4.
   Buy at 1 (day 6), sell at 4 (day 7), profit = 3.
   Total profit = 7.
   So the correct maximum profit is 7.

4. Final Answer:

   7 -> Option D

5. Quick Check:

   Two transactions max profit = 7 [OK]

Hint: Split array, find max profit left and right, sum max [OK]

Common Mistakes:

• Counting overlapping transactions
• Ignoring zero price days
• Assuming unlimited transactions